Answer
$\displaystyle C_{total} = \frac{2\kappa C_0}{\kappa +1}$
Work Step by Step
Because the thickness of the dielectric is not the same as the distance of the plate separation, we must treat this one capacitor as if it were two. One of these capacitors will have a plate separation distance the same length as the thickness of the dielectric and the other capacitor will have a plate separation distance of whatever distance is vacuum filled. These capacitors will be joined in series.
The original capacitor, without a dielectric in it, has a capacitace $C_0 = \displaystyle \frac{A\epsilon_0}{d}$
The vacuum-filled capacitor will have a capactice $C_{vacuum} = \displaystyle \frac{A\epsilon_0}{d/2} = 2\frac{A\epsilon_0}{d} = 2C_0$
The dielectric-filled capacitor will have a capacitance $C_{dielectric} = \displaystyle \kappa\frac{A\epsilon_0}{d/2} = 2\kappa\frac{A\epsilon_0}{d} = 2\kappa C_0$
-----Now we can solve:
$\displaystyle \frac{1}{C_{total}} = \frac{1}{C_{vacuum}} + \frac{1}{C_{dieletric}} = \frac{1}{2C_0} + \frac{1}{2\kappa C_0}$
$\displaystyle \frac{1}{C_{total}} = \frac{1}{2C_0}\left(1 + \frac{1}{\kappa}\right) = \frac{1}{2C_0}\left(\frac{\kappa + 1}{\kappa}\right)$
$\displaystyle \frac{1}{C_{total}} = \frac{\kappa + 1}{2\kappa C_0}$
$\displaystyle C_{total} = \frac{2\kappa C_0}{\kappa +1}$
